在三角形ABC中,若acos^2C/2+ccos^2A/2=3b/2,则求证:a+c=2b

a*cos^2(C/2)+c*cos^2(A/2)=3/2*b
a(cosC+1)/2+c(cosA+1)/2=1.5b
acosC+ccosA=3b-a-c
a*(a^2+b^2-c^2/2ab+c*（c^2+b^2-a^2）/2bc
=3b-a-c
2b^2/2b=3b-a-c
b=3b-a-c
a+c=2b

=============================================
[cos(C/2)]^2=(cosc+1)/2
所以原式即a*cosC+a+c*cosA+c=3b
而a*cosC+c*cosA=b (画张图看看就知道这个式子了)
于是化简为a+c=2b,
即得 证